JEE Main Study Notes for Quadratic Equation contain all the equations, properties, identities, and important formulas. The polynomial equations of the second degree in one variable is defined as a Quadratic Equation. Quadratic equation is a very important part of Algebra. In JEE Main Mathematics syllabus, quadratic equation is considered to be a difficult section.

• In JEE Main 2022, the candidates will be asked 2-3 questions from Quadratic Equation.
• JEE Main question paper has questions of around 8 marks from Quadratic Equation section.
• We have tried to explain The Sturm theorem, nature of roots, Formation of Quadratic Equations, The Wave curve Method, and some sample questions in this article.

Quick Links:

What is Quadratic Equation?

• The word ‘quadratic equation’ is derived from the Latin word ‘quadratus’ which means ‘a square’.
• A quadratic equation is an equation with the form, ax²+bx+c = 0, where x represents an unknown, a, b, and c are constants, and a is not equal to 0.
• If a = 0, then the equation becomes linear, not quadratic.
• The constants a, b, and c are known as the coefficients.
• ‘c’ expresses the constant term, b is the linear coefficient, and ‘a’ is the quadratic coefficient.
• The quadratic equations include only one unknown variable and hence, are considered to be univariate.
• As the greatest power is two, they are known as second-degree polynomial equations.

Discriminant of a Quadratic Equation

The discriminant of a quadratic equation is described as the number D= b²- 4ac and is found from the coefficients of the equation ax²+bx+c = 0. The discriminant shows the nature of roots that an equation has.

Note: b²- 4ac is found from the quadratic formula

The table given below lists the different kinds of roots that are associated with the values of the determinant.

For example, look at the quadratic equation y = 3x²+9x+5. Find its discriminant.

Solution: The quadratic equation given is y = 3x²+9x+5.

The formula of discriminant is D =b²- 4ac.

This means, a= 3, b= 9 and c= 5.

Therefore, the discriminant is,

D = 9²- 4.3.5 = 21.

Polynomial Equation of Degree n

A polynomial equation is an equation that can be expressed in the following form -

axⁿ + bx^{n - 1} + . . . + rx + s = 0,

where a, b, . . . , r and s are constants. The largest exponent of x that appears in a non-zero term of a polynomial is known as the degree of that polynomial.

For example,

1. Look at the equation 3x+1 = 0. The equation has degree 1 as the largest power of x which appears in the equation is 1. These equations are known as linear equations.
2. x²+ x - 3 = 0 has degree 2 because this is the largest power of x. Such degree 2 equations are known as quadratic equations or simply quadratics.
3. Degree 3 equations like x³ + 2x²- 4=0 are known as cubic.
4. A polynomial equation of degree n has n roots, but a few of them could be multiple roots. For example, consider x³-9x² +24x-16 =0.

It is quite obviously a polynomial of degree 3 and will therefore have three roots. The equation could be factored as (x-1) (x-4) (x-4) =0. This indicates that the roots of the equation are x=1, x=4, x=4. Hence, the root x=4 is repeated.

Must Read:

• JEE Main Study Notes for Matrices & Determinants
• JEE Main Study Notes for Mathematical Reasoning

Descartes Rule of Signs

Descartes’s Rule of Signs is a very popular rule that aids in getting an idea of the roots of a polynomial equation. This rule states that the number of positive real roots of Pn(x) = 0 cannot be higher than the number of significant changes. In a similar manner, the number of negative roots cannot be higher than the number of sign changes in Pn(-x).

For example, consider the equation P5(x) =x⁵+2x⁴+ x³+x²-x+2 = 0.

As given above, this equation has four sign changes. This means that it can have a maximum of four positive real roots. Now, it is seen that P5(-x) has only one sign change as,

P5(-x) = -x⁵+2x⁴+ x³+x²+x+2 = 0

Hence, it can hold only one negative real root.

Note: The Descartes Rule only provides an idea of the maximum number of positive or negative real roots but it does not give the exact number. To get an idea of the exact number of roots, Sturm’s Theorem must be used.

Check:

• JEE Main Mathematics Preparation Tips
• JEE Main Mathematics Question Paper

Sturm Theorem

Sturm theorem states that the number of real roots of the equation f(x) = 0 at [a, b] is equal to the difference between the number of sign changes in the Sturm sequence at x = a and x= b if f (a) ≠ 0 and f (b) ≠ 0.

It is known that any nth degree polynomial has exactly n roots. This makes it obvious that the number of complex roots is equal to the total roots – the number of real roots.

Hence, the number of complex roots = n - number of real roots, where a real root of multiplicity r has to be counted r times. In this case, the coefficients of the polynomial are real, the complex roots are α ± iβ, and therefore, the total number of complex roots is even.

The Wave Curve Method

Look at the question (x-3)(x-4)²(x-5)³=0.

The first step must be to make each factor equal to zero. Then, we get,

(x-3) = 0, (x-4)²= 0, (x-5)³= 0

• By equating all the terms to zero, various values of x which are known as the critical points are obtained. Here, 3, 4, and 5 are the critical points.
• Next, plot these points on a number line in increasing order. After plotting the points, enter a value higher than the rightmost point in the whole function.
• Here, the rightmost point is 5. Enter a value higher than 5, say, 6 in the whole function. When you put x=6 in the whole function, you get a positive sign so you can start drawing the curve above the number line.
• Now is the crucial step. Look for the exponents of every term. When the power of the factor is odd, you should change the side of the wavy curve (if it is above, then make it below and vice versa). Otherwise, you should continue with the same side of the number line.

Explanation: When the value 6 is entered, a positive sign is obtained. Draw the curve on the top side of the axis.

Now, 5 is touched. It is obtained from the factor(x-5)³. The exponent of this factor is odd. Therefore, the side of the curve must be changed. Again, for the next factor(x-4)², the power is even, and therefore, the side of the wavy curve must be retained. It was below the axis and will remain there only. The curve now looks like this –

The next factor is (x-3). Its power is 1 which is again odd. The side of the curve must be changed. At first, it was below, now it should be above the line.

Therefore, this is what the curve will look like. It is evident that the function is positive in the intervals when the curve rests above the axis and negative in the intervals where the curve rests below the axis.

Must Read:

• JEE Main Study Notes for Probability
• JEE Main Study Notes for Differential Calculus

Nature of Roots

The nature of roots of any equation depends upon the nature of its discriminant D.

1. If D < 0, there will be non-real complex roots, such roots are always conjugate to one another. I.e., if one root is p + iq then the other is p - iq, q ≠ 0.
2. If D = 0, then the roots are equal and real. Each root of the equation becomes -b/2a. This root is called repeated roots or double roots.
3. There will be real and unequal roots if D > 0.
4. D > 0 and D is a perfect square, if a, b, c are rational numbers. Then the roots of the equation are rational numbers and unequal.
5. If a, b, c are rational numbers, D > 0 but D is not a perfect square of a rational number, then the roots of the equation are irrational (surd). These types of roots are always conjugate to one another, i.e., if one root is p+√q, then the other one is p-√q, where q > 0.
6. If b and c are integers, a = 1, D > 0, and perfect square, then the roots of the equation are integers.

Also Check:

• JEE Main Study Notes for Algebra
• JEE Main Study notes for Integral Calculus

Relationship Between Roots and Coefficients

If a quadratic equation ax²+bx+c = 0 have roots namely α and β, then

1. α²+β²=(α+β)²-2αβ=b²-2ac/a²
2. α-β= √((α+β)² - 4αβ) = ±√b² -4ac/a = ±√D/a
3. α²-β²= (α+β) √((α+β)² - 4αβ) = -b√b² -4ac/a²
4. α³-β³= (α+β)³ +3αβ(α-β) = (b² -ac)√b² -4ac/a³
5. α³+β³= (α+β)³ -3αβ(α+β) = -b(b² -3ac)/a³
6. α² +αβ +β² = (α+β)² -αβ

Formation of Quadratic Equation with Given Roots

• An equation with roots α and β can be written as, (x – α) (x – β) = 0 or x²-(α+β)x+αβ=0 or x² - (sum of the roots)x + product of the roots = 0
• If a quadratic equation ax²+bx+c = 0 has roots α and β then, the identity will be ax²+bx+c = a(x- α) (x-β).

If you substitute the suitable value of x (real or imaginary) then, this identity will help to derive a number of relations between the roots.

Points to Remember on Quadratic Equation

• To be able to solve a quadratic equation of the form ax²+bx+c, the discriminant can be calculated using the formula D= b²- 4ac.
• The solution of the quadratic equation is expressed as x = [-b ± √ b² – 4ac] / 2a
• Suppose α and β are the roots of the quadratic equation ax²+bx+c = 0, then we will have the results given below for the sum and product of roots:

α + β = -b/a

α.β = c/a

α – β = √D/a

• It is impossible for a quadratic equation to hold three different roots and if, in any case, it occurs, then the equation turns into an identity.
• In the quadratic expression y =ax²+bx+c, where a, b, c ∈ R and a ≠ 0, the graph between x and y is usually a parabola.

1. When a > 0, the shape of the parabola is concave upwards.
2. When a < 0, the shape of the parabola is concave upwards.

• The inequalities of the form P(x)/ Q(x) > 0 could be easily solved using the method of intervals of number line rule.
• The maximum and minimum values of the form y = ax²+bx+c happen at the point x = -b/2a according to whether a > 0 or a< 0.

1. y ∈[(4ac-b²) / 4a, ∞] if a > 0
2. When a < 0, y ∈ [-∞, (4ac-b²) / 4a]

• The quadratic function of the form f(x, y) = ax2+by2 + 2hxy + 2gx + 2fy + c = 0 could be divided into two linear factors given that it satisfies the following condition: abc + 2fgh –af2 – bg2 – ch2 = 0
• Generally, if α₁ ,α₂ , α₃ , …… ,αn are the roots of the equation

f(x) = a₀xn +a₁xn^-1+ a₂xn^-2 + ……. + a_n-1x + a_n, then,

1. 1.Σα₁ = -a₁/a₀
2. 2.Σα₁α₂ = a₂/a₀
3. 3.Σα₁α₂α₃ = - a₃/a₀

……… ……….

Σ α₁α₂α₃…….α_n= (-1)n a_n/a₀

• Each equation of the nth degree consists of exactly n roots (n ≥1) and if it has more than n roots, then the equation turns into an identity.
• When there are two real numbers ‘a’ and ‘b’ such that f(a) and f(b) is of the opposing signs, then f(x) = 0 should have a minimum of one real root between ‘a’ and ‘b’.
• Each equation f(x) = 0 of the odd degree consists of a minimum of one real root of a sign which is opposite to that of its last term.

Check: JEE Main Study Noes for Sequence & Series

JEE Main Quadratic Equation - Solved Questions and Answers

Question: If p and q are the roots of the equation x²+px+q = 0, then

(a) p = 1, q = -2

(b) p = 0, q = 1

(c) p = -2, q = 0

(d) p = -2, q = 1

Answer: Option (a)

Solution: Given x²+px+q = 0

Sum of roots, p+q = -p

Product of roots, pq = q

=> p = 1

1+q = -1

=> q = -2

So p = 1, q = -2.

Question: Suppose the coefficient of x in the quadratic equation x²+ bx + c =0 was taken to be 17 instead of 13, its roots were found to be -2 and -15. Find the roots of the original quadratic equation.

Solution: As there is no change in the coefficient of x² and c, the product of zeros will stay the same for both the equations.

The product of zeros (c) = -2 × -15 = 30,

As the original value of b is 13.

∴ Sum of zeros = -b/a = -13.

Hence, the original quadratic equation is:

x²– (Sum of Zeros)x + (Product of Zeros) = 0

x² + 13x + 30 = 0

∴ (x + 10) (x + 3) = 0

Therefore, the roots of the original quadratic equations are -3 and -10.

Question: If one of the root of the equation x²+px+12 = 0 is 4, while the equation x²+px+q=0 has equal roots, then the value of q is

(a) 49/4

(b) 4

(c) 3

(d) 12

Answer: option (a)

Solution: Given x²+px+12 = 0 …(i)

Since 4 is a root of (i)

4²+4p+12 = 0

=> 4p = -28

=> p = -7

Given x²+px+q=0 has equal roots.

So D = 0

p²-4q = 0

=> 49-4q = 0

=> q = 49/4

Question: The value of a for which the sum of the squares of the roots of the equationx²-(a-2)x-a-1=0 assume the least value is

(a) 1

(b) 0

(c) 3

(d) 2

Solution:

Given x²-(a-2)x-a-1=0

Sum of roots, α + β = (a-2)

Product of roots, αβ = -a-1

Sum of squares of roots = α² + β² = (α+β)² -2αβ

= a²-2a+6

= (a-1)²+5

=> a = 1

Tricks to Solve Questions from Quadratic Equation

Question 1: Determine the values of k for which the quadratic expression (x – a) (x – 10) + 1 = 0 has integral roots.

Trick to solve this question: Rewrite the given equation as x² – (10 + k)x + 1 + 10k = 0. Determine the value of the discriminant and form an equation with k as (k-10)² – D = 4. Now, find the values of k.

Solution: k = 8 and 12.

Question 2: Determine the maximum or minimum value of the quadratic equation -4(x – 2)2 + 2.

Trick to solve this question: The value of a is negative. Therefore, the quadratic equation given is sure to have a maximum value. Now, find the maximum value.

Solution: 2

Question 3: How many real roots are there in the equation 2x⁵ + 2x⁴ – 11x³ + 9x² – 4x + 2 = 0?

Trick to solve this question: There are 4 sign changes in the given equation which means that it can have a maximum of 4 positive real roots. Note that for f(-x), there is only one sign change. Now, find the number of real roots in the equation.

Solution: The equation has only one negative real root.

Tips for Studying JEE Main Mathematics

1. Dedicate more effort and time to Algebra and Calculus because they have the highest weightage in the Mathematics section.
2. Thoroughly learn and revise all the formulae from calculus, probability, trigonometry, and geometry.
3. Make use of NCERT, Arihant, and R.D Sharma books for JEE Main Mathematics preparation.
4. Solve as many previous years’ question papers and mock tests as you can.

Quick Links: